The method for doing the 3x3x3 can be followed to do the 2x2x2. Since there are no edges to take care of the parity can be corrected from odd to even by a single turn of the U or D face, and the moves required calculated from there. Another idea is to do rotate U/D faces to make the number of corner moves required easier.
The parity in example 1 is even. In example 2 the parity is odd, and both ways of correcting it are shown.
Do the moves: R2U'FD'LU2L'DB'ULB'DL2FR'L'R2 on a solved cube. This is the same sequence that is used for the 3x3x3 example.
Starting with step 1 of the method, the twist of U is 0, so pairs to operate upon are in same layer (ie U or D).
orientate corners rotating pairs (21),(43),(56),(86)
(U')(LD2L'F'D2F)(U)-()(F'D2FLD2L')() for pair 1
(U)(LD2L'F'D2F)(U')-(U2)(F'D2FLD2L')(U2) for pair 2
Rotate cube forward 180 degrees so that U becomes D and vice versa, whilst L remains L and R remains R, plus F becomes B and vice versa.
()(LD2L'F'D2F)()-(U')(F'D2FLD2L')(U) for pair 3
(U)(LD2L'F'D2F)(U')-(U')(F'D2FLD2L')(U) for pair 4
Rotate cube forward 180 degrees so that U becomes D and vice versa, whilst L remains L and R remains R, plus F becomes B and vice versa. This undoes the cube rotation done earlier.
There are no edges so this step is skipped every time.
Writing the corners in terms of cycles we find the parity is even, and thus we skip step 3 of the method.
(1357)(2)(48)(6) has parity=3+0+1+0=4 which is even
We can now place corners as in step 4 of the method:
getting the corners to move in the U or D layer first
(DR2DR2-U2)(LF'LB2L'FLB2L2)(U2-R2D'R2D') changes (1357)(2)(48)(6) to (17)(2)(3)(48)(5)(6)
getting the corners to move in the U or D layer first
(UR2)
()(LD2L')()(U)(LD2L')(U')()(LD2L')() and
U2-()(LD2L')()(U)(LD2L')(U')()(LD2L')()-U2
(R2U')
In this example, the actual moves for orientating and moving the corners won't be explicitly given except where necesary, rather what you will be trying to achieve will be shown instead.
Do the moves: R2U'FD'LU2L'DB'ULB'DL2FR'L'R2 on a solved cube. Then do U. This is the same sequence as in example 1 followed by U.
For this cube configuration, the corners written in terms of cycles are:
(18457)(23)(6) which has parity 4+1+0=5
Simply doing U' changes the cube to the configuration given for example 1, so one possible solution is to do U' followed by the solution given in example 1.
Compare (18457)(23)(6) which has parity 4+1+0=5
with (1357)(2)(48)(6) which has parity=3+0+1+0=4.
The important question is how do you work out what the cycles will be after U'? The
simplest way is simply to recalculate having mentally done a U'.
Orient the corners using the method familiar from example 1.
Using any algorithm that swaps 2 corners but does nothing else to the 2x2x2 cube
will change the parity from odd to even and vice versa. On a 3x3x3 you would see the
edges affected in some way too.
Here I will use (U')FRU'R'URUR2F'RURU'R'(U) which swaps corners 2 and 3, thus reducing
(23) to (2)(3).
The cycles are now reduced to (18457)(2)(3)(6) which has parity 4+0+0+0=4 which is even,
and now the cycles can be solved using the method; this involves cycles (184)(157).
Thus the corners are moved by swapping a pair, and then doing two 3-cycles.